Why is $y^3-5y^2+6y-3=0$ if $x^3+2x^2+3x+3=0$ where $ y = α/(α+1) $?

Question

The beginning of a solution for a question in my book which is to find the value of $\sum α/(α+1)$ where α1, α2 and α3 are roots of $x^3+2x^2+3x+3=0$. For this in the solution they have said to take $ y = α/(α+1) $, and that therefore $y^3-5y^2+6y-3=0$. Where did this come from? How did they get this result? I tried substituting $ α = y/(1-y) $ but that just made it more complicated.

Answer 1

$$\begin{aligned}y^3-5y^2+6y-3&=\dfrac{\alpha^3}{(\alpha+1)^3}-\dfrac{5\alpha^2}{(\alpha+1)^2}+\dfrac{6\alpha}{\alpha+1}-3\\&=\dfrac{\alpha^3-5\alpha^2(\alpha+1)+6\alpha(\alpha+1)^2-3(\alpha+1)^3}{(\alpha+1)^3}\\ &=\dfrac{\alpha^3-5\alpha^3-5\alpha^2+6\alpha^3+12\alpha^2+6\alpha-3\alpha^3-9\alpha^2-9\alpha-3}{(\alpha+1)^3}\\ &=\dfrac{-\alpha^3-2\alpha^2-3\alpha-3}{(\alpha+1)^3}\\ &=\dfrac{-(\alpha^3+2\alpha^2+3\alpha+3)}{(\alpha+1)^3}\\ &=0 \end{aligned}$$ The last equality follows from the fact $\alpha$ is a root of $x^3+2x^2+3x+3$, so $\alpha^3+2\alpha^2+3\alpha+3=0$.

Answer 2

Putting $x=\dfrac y{1-y}$ in $$x^3+2x^2+3x+3=0$$

$$y^3+2(1-y)y^2+3y(1-y)^2+3(1-y)^3=0$$

$$\implies y^3(-3-3-1+1)+y^2(2-6+9)+y(\cdots)+3=0$$